https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Progamexd&feedformat=atom AoPS Wiki - User contributions [en] 2021-05-17T10:57:26Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_10&diff=93527 2015 AMC 10B Problems/Problem 10 2018-03-27T01:59:52Z <p>Progamexd: /* Solution */</p> <hr /> <div>==Problem==<br /> What are the sign and units digit of the product of all the odd negative integers strictly greater than &lt;math&gt;-2015&lt;/math&gt;?<br /> ==Solution==<br /> Since &lt;math&gt;-5&gt;-2015&lt;/math&gt;, the product must end with a &lt;math&gt;5&lt;/math&gt;.<br /> <br /> The multiplicands are the odd negative integers from &lt;math&gt;-1&lt;/math&gt; to &lt;math&gt;-2013&lt;/math&gt;. There are &lt;math&gt;\frac{|-2013+1|}2+1=1006+1&lt;/math&gt; of these numbers. Since &lt;math&gt;(-1)^{1007}=-1&lt;/math&gt;, the product is negative.<br /> <br /> Therefore, the answer must be &lt;math&gt;\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_10&diff=93526 2015 AMC 10B Problems/Problem 10 2018-03-27T01:56:34Z <p>Progamexd: /* Solution */</p> <hr /> <div>==Problem==<br /> What are the sign and units digit of the product of all the odd negative integers strictly greater than &lt;math&gt;-2015&lt;/math&gt;?<br /> ==Solution==<br /> Since &lt;math&gt;-5&gt;-2015&lt;/math&gt;, the product must end with a &lt;math&gt;5&lt;/math&gt;.<br /> <br /> The multiplicands are the odd negative integers from &lt;math&gt;-1&lt;/math&gt; to &lt;math&gt;-2013&lt;/math&gt;. There are &lt;math&gt;\frac{-2013+1}2+1=1006+1&lt;/math&gt; of these numbers. Since &lt;math&gt;(-1)^{1007}=-1&lt;/math&gt;, the product is negative.<br /> <br /> Therefore, the answer must be &lt;math&gt;\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_17&diff=87069 2012 AMC 10B Problems/Problem 17 2017-08-17T12:34:34Z <p>Progamexd: /* Solution */</p> <hr /> <div>==Problem==<br /> Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?<br /> <br /> &lt;math&gt;\mathbf{(A)}&lt;/math&gt; &lt;math&gt;\dfrac{1}{8}&lt;/math&gt; &lt;math&gt;\space&lt;/math&gt; &lt;math&gt;\mathbf{(B)}&lt;/math&gt; &lt;math&gt;\dfrac{1}{4}&lt;/math&gt; &lt;math&gt;\space&lt;/math&gt; &lt;math&gt;\mathbf{(C)}&lt;/math&gt; &lt;math&gt;\dfrac{\sqrt{10}}{10}&lt;/math&gt; &lt;math&gt;\space&lt;/math&gt; &lt;math&gt;\mathbf{(D)}&lt;/math&gt; &lt;math&gt;\dfrac{\sqrt{5}}{6}&lt;/math&gt; &lt;math&gt;\space&lt;/math&gt; &lt;math&gt;\mathbf{(E)}&lt;/math&gt; &lt;math&gt;\dfrac{\sqrt{5}}{5}&lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution==<br /> Let's find the volume of the smaller cone first. We know that the circumference of the paper disk is &lt;math&gt;24\pi&lt;/math&gt;, so the circumference of the smaller cone would be &lt;math&gt;\dfrac{120}{360} \times 24\pi = 8\pi&lt;/math&gt;. This means that the radius of the smaller cone is &lt;math&gt;4&lt;/math&gt;. Since the radius of the paper disk is &lt;math&gt;12&lt;/math&gt;, the slant height if the smaller cone would be &lt;math&gt;12&lt;/math&gt;. By the Pythagorean Theorem, the height of the cone is &lt;math&gt;\sqrt{12^2-4^2}=8\sqrt{2}&lt;/math&gt;. Thus, the volume of the smaller cone is &lt;math&gt;\dfrac{1}{3} \times 4^2\pi \times 8\sqrt{2}=\dfrac{128\sqrt{2}}{3}\pi&lt;/math&gt;.<br /> <br /> Now, we need to find the volume of the larger cone. Using the same reason as above, we get that the radius is &lt;math&gt;8&lt;/math&gt; and the slant height is &lt;math&gt;12&lt;/math&gt;. By the Pythagorean Theorem again, the height is &lt;math&gt;\sqrt{12^2-8^2}=4\sqrt{5}&lt;/math&gt;. Thus, the volume of the larger cone is &lt;math&gt;\dfrac{1}{3} \times 8^2\pi \times 4\sqrt{5} = \dfrac{256\sqrt{5}}{3}\pi&lt;/math&gt;.<br /> <br /> The question asked for the ratio of the volume of the smaller cone to the larger cone. We need to find &lt;math&gt;\dfrac{\dfrac{128\sqrt{2}}{3}\pi}{\dfrac{256\sqrt{5}}{3}\pi}=\dfrac{\sqrt{10}}{10}&lt;/math&gt; after simplifying, or &lt;math&gt;\boxed{C}&lt;/math&gt; .<br /> <br /> ~A side note<br /> <br /> We can first simplify the volume ratio: &lt;math&gt;\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.&lt;/math&gt; Now we can find the GENERAL formulas for &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;h&lt;/math&gt; based on the original circle radius and angle to cut out, then we can substitute the appropriate numbers, which gives us &lt;math&gt;C.&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_17&diff=87068 2012 AMC 10B Problems/Problem 17 2017-08-17T12:34:14Z <p>Progamexd: /* Faster Solution */</p> <hr /> <div>==Problem==<br /> Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?<br /> <br /> &lt;math&gt;\mathbf{(A)}&lt;/math&gt; &lt;math&gt;\dfrac{1}{8}&lt;/math&gt; &lt;math&gt;\space&lt;/math&gt; &lt;math&gt;\mathbf{(B)}&lt;/math&gt; &lt;math&gt;\dfrac{1}{4}&lt;/math&gt; &lt;math&gt;\space&lt;/math&gt; &lt;math&gt;\mathbf{(C)}&lt;/math&gt; &lt;math&gt;\dfrac{\sqrt{10}}{10}&lt;/math&gt; &lt;math&gt;\space&lt;/math&gt; &lt;math&gt;\mathbf{(D)}&lt;/math&gt; &lt;math&gt;\dfrac{\sqrt{5}}{6}&lt;/math&gt; &lt;math&gt;\space&lt;/math&gt; &lt;math&gt;\mathbf{(E)}&lt;/math&gt; &lt;math&gt;\dfrac{\sqrt{5}}{5}&lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution==<br /> Let's find the volume of the smaller cone first. We know that the circumference of the paper disk is &lt;math&gt;24\pi&lt;/math&gt;, so the circumference of the smaller cone would be &lt;math&gt;\dfrac{120}{360} \times 24\pi = 8\pi&lt;/math&gt;. This means that the radius of the smaller cone is &lt;math&gt;4&lt;/math&gt;. Since the radius of the paper disk is &lt;math&gt;12&lt;/math&gt;, the slant height if the smaller cone would be &lt;math&gt;12&lt;/math&gt;. By the Pythagorean Theorem, the height of the cone is &lt;math&gt;\sqrt{12^2-4^2}=8\sqrt{2}&lt;/math&gt;. Thus, the volume of the smaller cone is &lt;math&gt;\dfrac{1}{3} \times 4^2\pi \times 8\sqrt{2}=\dfrac{128\sqrt{2}}{3}\pi&lt;/math&gt;.<br /> <br /> Now, we need to find the volume of the larger cone. Using the same reason as above, we get that the radius is &lt;math&gt;8&lt;/math&gt; and the slant height is &lt;math&gt;12&lt;/math&gt;. By the Pythagorean Theorem again, the height is &lt;math&gt;\sqrt{12^2-8^2}=4\sqrt{5}&lt;/math&gt;. Thus, the volume of the larger cone is &lt;math&gt;\dfrac{1}{3} \times 8^2\pi \times 4\sqrt{5} = \dfrac{256\sqrt{5}}{3}\pi&lt;/math&gt;.<br /> <br /> The question asked for the ratio of the volume of the smaller cone to the larger cone. We need to find &lt;math&gt;\dfrac{\dfrac{128\sqrt{2}}{3}\pi}{\dfrac{256\sqrt{5}}{3}\pi}=\dfrac{\sqrt{10}}{10}&lt;/math&gt; after simplifying, or &lt;math&gt;\boxed{C}&lt;/math&gt; .<br /> <br /> ~A side note<br /> We can first simplify the volume ratio: &lt;math&gt;\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.&lt;/math&gt; Now we can find the GENERAL formulas for &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;h&lt;/math&gt; based on the original circle radius and angle to cut out, then we can substitute the appropriate numbers, which gives us &lt;math&gt;C.&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_19&diff=87025 2011 AMC 10B Problems/Problem 19 2017-08-14T16:24:56Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>== Problem==<br /> <br /> What is the product of all the roots of the equation &lt;cmath&gt;\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.&lt;/cmath&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> First, square both sides, and isolate the absolute value.<br /> &lt;cmath&gt;\begin{align*}<br /> 5|x|+8&amp;=x^2-16\\<br /> 5|x|&amp;=x^2-24\\<br /> |x|&amp;=\frac{x^2-24}{5}<br /> \end{align*}&lt;/cmath&gt;<br /> Solve for the absolute value and factor.<br /> &lt;cmath&gt;x=\frac{x^2-24}{5} \longrightarrow 5x=x^2-24 \longrightarrow 0=x^2-5x-24 \longrightarrow (x-8)(x+3)=0\\<br /> x=\frac{24-x^2}{5} \longrightarrow 5x=24-x^2 \longrightarrow 0=x^2+5x-24 \longrightarrow (x+8)(x-3)=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x= -8, -3, 3, 8&lt;/cmath&gt;<br /> <br /> However, this is not the final answer. Plug it back into the original equation to ensure it still works. Whether the number is positive or negative does not matter since the absolute value or square will cancel it out anyways.<br /> &lt;cmath&gt;\sqrt{5|3|+8}=\sqrt{3^2-16} \longrightarrow\sqrt{15+8}=\sqrt{9-16} \longrightarrow \sqrt{23}\not=\sqrt{-7}\\<br /> \sqrt{5|8|+8}=\sqrt{8^2-16} \longrightarrow \sqrt{40+8}=\sqrt{64-16} \longrightarrow \sqrt{48}=\sqrt{48}&lt;/cmath&gt;<br /> The roots of this equation are &lt;math&gt;-8&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; and product is &lt;math&gt;-8 \times 8 = \boxed{\textbf{(A)} -64}&lt;/math&gt;<br /> <br /> == Solution 2 == <br /> <br /> Square both sides, to get &lt;math&gt;5|x| + 8 = x^2-16&lt;/math&gt;. Rearrange to get &lt;math&gt;x^2 - 5|x| - 24 = 0&lt;/math&gt;. Seeing that &lt;math&gt;x^2 = |x|^2&lt;/math&gt;, substitute to get &lt;math&gt;|x|^2 - 5|x| - 24 = 0&lt;/math&gt;. We see that this is a quadratic in |x|. Factoring, we get &lt;math&gt;(|x|-8)(|x|+3) = 0&lt;/math&gt;, so &lt;math&gt;|x| = \{8,-3\}&lt;/math&gt;. Since the radicand of the equation can't be negative, the sole solution is &lt;math&gt;|x| = 8&lt;/math&gt;. Therefore, the x can be &lt;math&gt;8&lt;/math&gt; or &lt;math&gt;-8&lt;/math&gt;. The product is then &lt;math&gt;-8 \times 8 = \boxed{\textbf{(A)} -64}&lt;/math&gt;.<br /> <br /> == See Also==<br /> <br /> {{AMC10 box|year=2011|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_3&diff=86969 2010 AMC 10A Problems/Problem 3 2017-08-10T12:30:57Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>== Problem 3 ==<br /> Tyrone had &lt;math&gt;97&lt;/math&gt; marbles and Eric had &lt;math&gt;11&lt;/math&gt; marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 3<br /> \qquad<br /> \mathrm{(B)}\ 13<br /> \qquad<br /> \mathrm{(C)}\ 18<br /> \qquad<br /> \mathrm{(D)}\ 25<br /> \qquad<br /> \mathrm{(E)}\ 29<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; be the number of marbles Tyrone gave to Eric. Then, &lt;math&gt;97-x = 2\cdot(11+x)&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; yields &lt;math&gt;75=3x&lt;/math&gt; and &lt;math&gt;x = 25&lt;/math&gt;. The answer is &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2 ==<br /> Since the amount of balls Tyler and Eric have a specific ratio when Tyler give some of his balls to Eric, we can divide the balls to their respective ratios. Altogether, they have &lt;math&gt;97+11&lt;/math&gt; balls, or &lt;math&gt;108&lt;/math&gt; balls, and it does not change when Tyler give some of his to Eric, since none are lost in the process. After Tyler give a few of his balls away, the ratio between the balls is &lt;math&gt;2:1&lt;/math&gt;, meaning that together added up is &lt;math&gt;108&lt;/math&gt;. The two numbers add up to &lt;math&gt;3&lt;/math&gt; and we divide &lt;math&gt;108&lt;/math&gt; by &lt;math&gt;3&lt;/math&gt;, and outcomes &lt;math&gt;36&lt;/math&gt;. This is the amount of balls Eric has, and so doubling that results in the amount of balls Tyler has, which is &lt;math&gt;72&lt;/math&gt;. &lt;math&gt;97-72&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, and &lt;math&gt;36-11&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, thus proving our statement true, and the answer is &lt;math&gt;\boxed{D=25}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_3&diff=86968 2010 AMC 10A Problems/Problem 3 2017-08-10T12:30:23Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>== Problem 3 ==<br /> Tyrone had &lt;math&gt;97&lt;/math&gt; marbles and Eric had &lt;math&gt;11&lt;/math&gt; marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 3<br /> \qquad<br /> \mathrm{(B)}\ 13<br /> \qquad<br /> \mathrm{(C)}\ 18<br /> \qquad<br /> \mathrm{(D)}\ 25<br /> \qquad<br /> \mathrm{(E)}\ 29<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; be the number of marbles Tyrone gave to Eric. Then, &lt;math&gt;97-x = 2\cdot(11+x)&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; yields &lt;math&gt;75=3x&lt;/math&gt; and &lt;math&gt;x = 25&lt;/math&gt;. The answer is &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2 ==<br /> Since the amount of balls Tyler and Eric have a specific ratio when Tyler give some of his balls to Eric, we can divide the balls to their respective ratios. Altogether, they have &lt;math&gt;97+11&lt;/math&gt; balls, or &lt;math&gt;108&lt;/math&gt; balls, and it does not change when Tyler give some of his to Eric, since none are lost in the process. After Tyler give a few of his balls away, the ratio between the balls is &lt;math&gt;2:1&lt;/math&gt;, meaning that together added up is &lt;math&gt;108&lt;/math&gt;. The two numbers add up to &lt;math&gt;3&lt;/math&gt; and we divide &lt;math&gt;108&lt;/math&gt; by &lt;math&gt;3&lt;/math&gt;, and outcomes &lt;math&gt;36&lt;/math&gt;. This is the amount of balls Eric has, and so doubling that results in the amount of balls Tyler has, which is &lt;math&gt;72&lt;/math&gt;. &lt;math&gt;97-72&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, and &lt;math&gt;36-11&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, thus proving our statement true, and the answer is &lt;math&gt;25&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_3&diff=86967 2010 AMC 10A Problems/Problem 3 2017-08-10T12:30:00Z <p>Progamexd: /* Solution */</p> <hr /> <div>== Problem 3 ==<br /> Tyrone had &lt;math&gt;97&lt;/math&gt; marbles and Eric had &lt;math&gt;11&lt;/math&gt; marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 3<br /> \qquad<br /> \mathrm{(B)}\ 13<br /> \qquad<br /> \mathrm{(C)}\ 18<br /> \qquad<br /> \mathrm{(D)}\ 25<br /> \qquad<br /> \mathrm{(E)}\ 29<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; be the number of marbles Tyrone gave to Eric. Then, &lt;math&gt;97-x = 2\cdot(11+x)&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; yields &lt;math&gt;75=3x&lt;/math&gt; and &lt;math&gt;x = 25&lt;/math&gt;. The answer is &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2 ==<br /> Since the amount of balls Tyler and Eric have a specific ratio when Tyler give some of his balls to Eric, we can divide the balls to their respective ratios. Altogether, they have &lt;math&gt;97+11&lt;/math&gt; balls, or &lt;math&gt;108&lt;/math&gt; balls, and it does not change when Tyler give some of his to Eric, since none are lost in the process. After Tyler give a few of his balls away, the ratio between the balls is &lt;math&gt;2:1&lt;/math&gt;, meaning that together added up is &lt;math&gt;108&lt;/math&gt;. The two numbers add up to &lt;math&gt;3&lt;/math&gt; and we divide &lt;math&gt;108&lt;/math&gt; by &lt;math&gt;3&lt;/math&gt;, and outcomes &lt;math&gt;36&lt;/math&gt;. This is the amount of balls Eric has, and so doubling that results in the amount of balls Tyler has, which is &lt;math&gt;72&lt;/math&gt;. &lt;math&gt;97-72&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, and &lt;math&gt;36-11&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, thus proving our statement true.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_20&diff=86819 2017 AMC 10A Problems/Problem 20 2017-08-05T14:04:08Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; equal the sum of the digits of positive integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;S(1507) = 13&lt;/math&gt;. For a particular positive integer &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S(n) = 1274&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;S(n+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;n \equiv S(n) \pmod{9}&lt;/math&gt;. This can be seen from the fact that &lt;math&gt;\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}&lt;/math&gt;. Thus, if &lt;math&gt;S(n) = 1274&lt;/math&gt;, then &lt;math&gt;n \equiv 5 \pmod{9}&lt;/math&gt;, and thus &lt;math&gt;n+1 \equiv S(n+1) \equiv 6 \pmod{9}&lt;/math&gt;. The only answer choice that is &lt;math&gt;6 \pmod{9}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> One divisibility rule for division that we can use in the problem is that a multiple of &lt;math&gt;9&lt;/math&gt; has its digit always add up to a multiple of &lt;math&gt;9&lt;/math&gt;. We can find out that the least number of digits the number &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;142&lt;/math&gt;, with &lt;math&gt;141&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;'s and &lt;math&gt;1&lt;/math&gt; &lt;math&gt;5&lt;/math&gt;, assuming the rule above. No matter what arrangement or different digits we use, the divisor rule stays the same. To make the problem simpler, we can just use the &lt;math&gt;141&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;'s and &lt;math&gt;1&lt;/math&gt; &lt;math&gt;5&lt;/math&gt;. By randomly mixing the digits up, we are likely to get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9995999&lt;/math&gt;...&lt;math&gt;9999&lt;/math&gt;. By adding &lt;math&gt;1&lt;/math&gt; to this number, we get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9996000&lt;/math&gt;...&lt;math&gt;0000&lt;/math&gt;.<br /> Knowing that this number is ONLY divisible by &lt;math&gt;9&lt;/math&gt; when &lt;math&gt;6&lt;/math&gt; is subtracted, we can subtract &lt;math&gt;6&lt;/math&gt; from every available choice, and see if the number is divisible by &lt;math&gt;9&lt;/math&gt; afterwards.<br /> After subtracting &lt;math&gt;6&lt;/math&gt; from every number, we can conclude that &lt;math&gt;1233&lt;/math&gt; (originally &lt;math&gt;1239&lt;/math&gt;) is the only number divisible by &lt;math&gt;9&lt;/math&gt;. <br /> So our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> The number &lt;math&gt;n&lt;/math&gt; can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice.<br /> <br /> If &lt;math&gt;\boxed{\textbf{(A)}\ 1}&lt;/math&gt; is correct, then &lt;math&gt;n&lt;/math&gt; must be some number &lt;math&gt;99999999...9&lt;/math&gt;, because when we add one to &lt;math&gt;99999999...9&lt;/math&gt; we get &lt;math&gt;10000000...00&lt;/math&gt;. Thus, if &lt;math&gt;1&lt;/math&gt; is the correct answer, then the equation &lt;math&gt;9x=1274&lt;/math&gt; must have an integer solution (i.e. &lt;math&gt;1274&lt;/math&gt; must be divisible by &lt;math&gt;9&lt;/math&gt;). But since it does not, &lt;math&gt;1&lt;/math&gt; is not the correct answer.<br /> <br /> If &lt;math&gt;\boxed{\textbf{(B)}\ 3}&lt;/math&gt; is correct, then &lt;math&gt;n&lt;/math&gt; must be some number &lt;math&gt;29999999...9&lt;/math&gt;, because when we add one to &lt;math&gt;29999999...9&lt;/math&gt;, we get &lt;math&gt;30000000...00&lt;/math&gt;. Thus, if &lt;math&gt;2&lt;/math&gt; is the correct answer, then the equation &lt;math&gt;2+9x=1274&lt;/math&gt; must have an integer solution. But since it does not, &lt;math&gt;3&lt;/math&gt; is not the correct answer.<br /> <br /> Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities. Notice that if &lt;math&gt;S(n+1)=N&lt;/math&gt;, then &lt;math&gt;n&lt;/math&gt; must be a number whose initial digits sum to &lt;math&gt;N-1&lt;/math&gt;, and whose other, terminating digits, are all &lt;math&gt;9&lt;/math&gt;. Thus, we can evaluate the three final possibilities by seeing if the equation &lt;math&gt;(N-1)+9x=1274&lt;/math&gt; has an integer solution.<br /> <br /> The equation does not have an integer solution for &lt;math&gt;N=12&lt;/math&gt;, so &lt;math&gt;\boxed{\textbf{(C)}\ 12}&lt;/math&gt; is not correct. However, the equation does have an integer solution for &lt;math&gt;N=1239&lt;/math&gt; (&lt;math&gt;x=4&lt;/math&gt;), so &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt; is the answer.<br /> <br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_19&diff=86548 2007 AMC 10A Problems/Problem 19 2017-07-24T12:33:24Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width?<br /> &lt;asy&gt;<br /> unitsize(3 cm);<br /> <br /> real w = (sqrt(2) - 1)/2;<br /> <br /> filldraw((0,0)--(w*sqrt(2)/2,0)--(0.5,0.5 - w*sqrt(2)/2)--(1 - w*sqrt(2)/2,0)--(1,0)--(1,w*sqrt(2)/2)--(0.5 + w*sqrt(2)/2,0.5)--(1,1 - w*sqrt(2)/2)--(1,1)--(1 - w*sqrt(2)/2,1)--(0.5,0.5 + w*sqrt(2)/2)--(w*sqrt(2)/2,1)--(0,1)--(0,1 - w*sqrt(2)/2)--(0.5 - w*sqrt(2)/2,0.5)--(0,w*sqrt(2)/2)--cycle,gray(0.7));<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\text{(A)}\ 2\sqrt {2} + 1 \qquad \text{(B)}\ 3\sqrt {2}\qquad \text{(C)}\ 2\sqrt {2} + 2 \qquad \text{(D)}\ 3\sqrt {2} + 1 \qquad \text{(E)}\ 3\sqrt {2} + 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> [[Without loss of generality]], let the side length of the square be &lt;math&gt;1&lt;/math&gt; unit. The area of the painted area is &lt;math&gt;\frac{1}2&lt;/math&gt; of the area of the larger square, so the total unpainted area is also &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> Each of the &lt;math&gt;4&lt;/math&gt; unpainted triangle has area &lt;math&gt;\frac{1}8&lt;/math&gt;. It is easy to tell that these triangles are isosceles right triangles, so let &lt;math&gt;a&lt;/math&gt; be the side length of one of the smaller triangles:<br /> &lt;center&gt;<br /> &lt;math&gt;\frac{a^2}2 = \frac{1}8 \rightarrow a= \frac{1}2&lt;/math&gt;<br /> &lt;/center&gt;<br /> The diagonal of the triangle is &lt;math&gt;\frac{\sqrt{2}}2&lt;/math&gt;. The corners of the painted areas are also isosceles right triangles with side length &lt;math&gt;\frac{1-\frac{\sqrt{2}}2}2 = \frac{1}2-\frac{\sqrt2}4&lt;/math&gt;. Its diagonal is equal to the width of the paint, and is &lt;math&gt;\frac{\sqrt{2}}2-\frac{1}2&lt;/math&gt;. The answer we are looking for is thus &lt;math&gt;\frac{1}{\frac{\sqrt{2}}2-\frac{1}2}&lt;/math&gt;. Multiply the numerator and the denominator by &lt;math&gt;\frac{\sqrt{2}}2+\frac{1}2&lt;/math&gt; to simplify, and you get &lt;math&gt;\frac{\frac{\sqrt{2}}{2}+\frac{1}{2}}{\frac{2}{4}-\frac{1}{4}}&lt;/math&gt; or &lt;math&gt;4\left(\frac{\sqrt{2}}{2}+\frac{1}{2}\right)&lt;/math&gt; which is &lt;math&gt;\boxed{2\sqrt{2}+2} \rightarrow C&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Again, have the length of the square equal to &lt;math&gt;1&lt;/math&gt; and let the width of each individual stripe be &lt;math&gt;n&lt;/math&gt;. Note that you can split each stripe into two rectangles and two isosceles right triangles at the corners. Then the area of each stripe is &lt;math&gt;n(\sqrt{2}-\frac{n}{2})=\sqrt{2}n-\frac{n^2}{2}&lt;/math&gt;. The area covered by the two total stripes is twice the area of one stripe, minus the area in the intersection of the stripes, which is a square with side length &lt;math&gt;n&lt;/math&gt;. This area is equal to &lt;math&gt;\frac{1}{2}&lt;/math&gt; So:<br /> &lt;center&gt;<br /> &lt;math&gt;2\sqrt2n-2n^2=\frac{1}{2}\rightarrow-2n^2+2\sqrt2n-\frac{1}{2}=0&lt;/math&gt;.<br /> &lt;/center&gt;<br /> By the [[quadratic formula]], <br /> &lt;center&gt;<br /> &lt;math&gt;n=\frac{-2\sqrt{2}\pm\sqrt{(2\sqrt{2})^2-4(-2)(-\frac{1}{2})}}{2(-2)}\rightarrow n=\frac{-2\sqrt{2}\pm 2}{-4}\rightarrow n=\frac{\sqrt{2}\pm 1}{2}&lt;/math&gt;<br /> &lt;/center&gt;<br /> It's easy to tell that &lt;math&gt;\frac{\sqrt{2}+1}{2}&lt;/math&gt; is too large, so &lt;math&gt;n=\frac{\sqrt{2}-1}{2}&lt;/math&gt;. We want to find &lt;math&gt;\frac{1}{n}&lt;/math&gt;, and &lt;math&gt;\frac{1}{n}=\frac{2}{\sqrt{2}-1}&lt;/math&gt;. Multiply the numerator and the denominator by &lt;math&gt;\sqrt{2}+1&lt;/math&gt;, <br /> &lt;center&gt;<br /> &lt;math&gt;\frac{1}{n}=\frac{2(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=\boxed{2\sqrt{2}+2}\rightarrow C&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2007|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_19&diff=86547 2007 AMC 10A Problems/Problem 19 2017-07-24T12:33:09Z <p>Progamexd: /* Solution */</p> <hr /> <div>==Problem==<br /> A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width?<br /> &lt;asy&gt;<br /> unitsize(3 cm);<br /> <br /> real w = (sqrt(2) - 1)/2;<br /> <br /> filldraw((0,0)--(w*sqrt(2)/2,0)--(0.5,0.5 - w*sqrt(2)/2)--(1 - w*sqrt(2)/2,0)--(1,0)--(1,w*sqrt(2)/2)--(0.5 + w*sqrt(2)/2,0.5)--(1,1 - w*sqrt(2)/2)--(1,1)--(1 - w*sqrt(2)/2,1)--(0.5,0.5 + w*sqrt(2)/2)--(w*sqrt(2)/2,1)--(0,1)--(0,1 - w*sqrt(2)/2)--(0.5 - w*sqrt(2)/2,0.5)--(0,w*sqrt(2)/2)--cycle,gray(0.7));<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\text{(A)}\ 2\sqrt {2} + 1 \qquad \text{(B)}\ 3\sqrt {2}\qquad \text{(C)}\ 2\sqrt {2} + 2 \qquad \text{(D)}\ 3\sqrt {2} + 1 \qquad \text{(E)}\ 3\sqrt {2} + 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> [[Without loss of generality]], let the side length of the square be &lt;math&gt;1&lt;/math&gt; unit. The area of the painted area is &lt;math&gt;\frac{1}2&lt;/math&gt; of the area of the larger square, so the total unpainted area is also &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> Each of the &lt;math&gt;4&lt;/math&gt; unpainted triangle has area &lt;math&gt;\frac{1}8&lt;/math&gt;. It is easy to tell that these triangles are isosceles right triangles, so let &lt;math&gt;a&lt;/math&gt; be the side length of one of the smaller triangles:<br /> &lt;center&gt;<br /> &lt;math&gt;\frac{a^2}2 = \frac{1}8 \rightarrow a= \frac{1}2&lt;/math&gt;<br /> &lt;/center&gt;<br /> The diagonal of the triangle is &lt;math&gt;\frac{\sqrt{2}}2&lt;/math&gt;. The corners of the painted areas are also isosceles right triangles with side length &lt;math&gt;\frac{1-\frac{\sqrt{2}}2}2 = \frac{1}2-\frac{\sqrt2}4&lt;/math&gt;. Its diagonal is equal to the width of the paint, and is &lt;math&gt;\frac{\sqrt{2}}2-\frac{1}2&lt;/math&gt;. The answer we are looking for is thus &lt;math&gt;\frac{1}{\frac{\sqrt{2}}2-\frac{1}2}&lt;/math&gt;. Multiply the numerator and the denominator by &lt;math&gt;\frac{\sqrt{2}}2+\frac{1}2&lt;/math&gt; to simplify, and you get &lt;math&gt;\frac{\frac{\sqrt{2}}{2}+\frac{1}{2}}{\frac{2}{4}-\frac{1}{4}}&lt;/math&gt; or &lt;math&gt;4\left(\frac{\sqrt{2}}{2}+\frac{1}{2}\right)&lt;/math&gt; which is &lt;math&gt;\boxed{2\sqrt{2}+2} \rightarrow C&lt;/math&gt;.<br /> <br /> =Solution 2=<br /> <br /> Again, have the length of the square equal to &lt;math&gt;1&lt;/math&gt; and let the width of each individual stripe be &lt;math&gt;n&lt;/math&gt;. Note that you can split each stripe into two rectangles and two isosceles right triangles at the corners. Then the area of each stripe is &lt;math&gt;n(\sqrt{2}-\frac{n}{2})=\sqrt{2}n-\frac{n^2}{2}&lt;/math&gt;. The area covered by the two total stripes is twice the area of one stripe, minus the area in the intersection of the stripes, which is a square with side length &lt;math&gt;n&lt;/math&gt;. This area is equal to &lt;math&gt;\frac{1}{2}&lt;/math&gt; So:<br /> &lt;center&gt;<br /> &lt;math&gt;2\sqrt2n-2n^2=\frac{1}{2}\rightarrow-2n^2+2\sqrt2n-\frac{1}{2}=0&lt;/math&gt;.<br /> &lt;/center&gt;<br /> By the [[quadratic formula]], <br /> &lt;center&gt;<br /> &lt;math&gt;n=\frac{-2\sqrt{2}\pm\sqrt{(2\sqrt{2})^2-4(-2)(-\frac{1}{2})}}{2(-2)}\rightarrow n=\frac{-2\sqrt{2}\pm 2}{-4}\rightarrow n=\frac{\sqrt{2}\pm 1}{2}&lt;/math&gt;<br /> &lt;/center&gt;<br /> It's easy to tell that &lt;math&gt;\frac{\sqrt{2}+1}{2}&lt;/math&gt; is too large, so &lt;math&gt;n=\frac{\sqrt{2}-1}{2}&lt;/math&gt;. We want to find &lt;math&gt;\frac{1}{n}&lt;/math&gt;, and &lt;math&gt;\frac{1}{n}=\frac{2}{\sqrt{2}-1}&lt;/math&gt;. Multiply the numerator and the denominator by &lt;math&gt;\sqrt{2}+1&lt;/math&gt;, <br /> &lt;center&gt;<br /> &lt;math&gt;\frac{1}{n}=\frac{2(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=\boxed{2\sqrt{2}+2}\rightarrow C&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2007|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_9&diff=86546 2007 AMC 12A Problems/Problem 9 2017-07-24T12:28:23Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #9]] and [[2007 AMC 10A Problems/Problem 13|2007 AMC 10A #13]]}}<br /> ==Problem==<br /> Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the [[ratio]] of Yan's distance from his home to his distance from the stadium?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the distance from Yan's initial position to the stadium be &lt;math&gt;a&lt;/math&gt; and the distance from Yan's initial position to home be &lt;math&gt;b&lt;/math&gt;. We are trying to find &lt;math&gt;b/a&lt;/math&gt;, and we have the following identity given by the problem:<br /> <br /> &lt;cmath&gt;\begin{align*}a &amp;= b + \frac{a+b}{7}\\<br /> \frac{6a}{7} &amp;= \frac{8b}{7} \Longrightarrow 6a = 8b\end{align*}&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;b/a = 6/8 = 3/4&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(B)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Another way of solving this problem is by setting the distance between Yan's home and the stadium, thus filling in one variable.<br /> Let us set the distance between the two places to be &lt;math&gt;42z&lt;/math&gt;, where &lt;math&gt;z&lt;/math&gt; is a random measurement (cause life, why not?) The distance to going to his home then riding his bike, which is &lt;math&gt;7&lt;/math&gt; times faster, is equal to him just walking to the stadium. So the equation would be:<br /> Let &lt;math&gt;x=&lt;/math&gt; the distance from Yan's position to his home. <br /> Let &lt;math&gt;42z=&lt;/math&gt; the distance from Yan's home to the stadium.<br /> <br /> &lt;math&gt;42-x=x+42/7&lt;/math&gt;<br /> <br /> &lt;math&gt;42-x=x+6&lt;/math&gt;<br /> <br /> &lt;math&gt;36=2x&lt;/math&gt;<br /> <br /> &lt;math&gt;x=18&lt;/math&gt;<br /> <br /> But we're still not done with the question. We know that Yan is &lt;math&gt;18z&lt;/math&gt; from his home, and is &lt;math&gt;42z-18z&lt;/math&gt; or &lt;math&gt;24z&lt;/math&gt; from the stadium.<br /> &lt;math&gt;18z/24z&lt;/math&gt;, the &lt;math&gt;z&lt;/math&gt;'s cancel out, and we are left with &lt;math&gt;3/4&lt;/math&gt;.<br /> Thus, the answer is &lt;math&gt;\mathrm{(B)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> ~ProGameXD<br /> <br /> ==See also==<br /> {{AMC12 box|year=2007|ab=A|num-b=8|num-a=10}}<br /> {{AMC10 box|year=2007|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_9&diff=86545 2007 AMC 12A Problems/Problem 9 2017-07-24T12:27:47Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #9]] and [[2007 AMC 10A Problems/Problem 13|2007 AMC 10A #13]]}}<br /> ==Problem==<br /> Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the [[ratio]] of Yan's distance from his home to his distance from the stadium?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the distance from Yan's initial position to the stadium be &lt;math&gt;a&lt;/math&gt; and the distance from Yan's initial position to home be &lt;math&gt;b&lt;/math&gt;. We are trying to find &lt;math&gt;b/a&lt;/math&gt;, and we have the following identity given by the problem:<br /> <br /> &lt;cmath&gt;\begin{align*}a &amp;= b + \frac{a+b}{7}\\<br /> \frac{6a}{7} &amp;= \frac{8b}{7} \Longrightarrow 6a = 8b\end{align*}&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;b/a = 6/8 = 3/4&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(B)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Another way of solving this problem is by setting the distance between Yan's home and the stadium, thus filling in one variable.<br /> Let us set the distance between the two places to be &lt;math&gt;42z&lt;/math&gt;, where &lt;math&gt;z&lt;/math&gt; is a random measurement (cause life, why not?). The distance to going to his home then riding his bike, which is &lt;math&gt;7&lt;/math&gt; times faster, is equal to him just walking to the stadium. So the equation would be:<br /> Let &lt;math&gt;x=&lt;/math&gt; the distance from Yan's position to his home. <br /> Let &lt;math&gt;42z=&lt;/math&gt; the distance from Yan's home to the stadium.<br /> <br /> &lt;math&gt;42-x=x+42/7&lt;/math&gt;<br /> <br /> &lt;math&gt;42-x=x+6&lt;/math&gt;<br /> <br /> &lt;math&gt;36=2x&lt;/math&gt;<br /> <br /> &lt;math&gt;x=18&lt;/math&gt;<br /> <br /> But we're still not done with the question. We know that Yan is &lt;math&gt;18z&lt;/math&gt; from his home, and is &lt;math&gt;42z-18z&lt;/math&gt; or &lt;math&gt;24z&lt;/math&gt; from the stadium.<br /> &lt;math&gt;18z/24z&lt;/math&gt;, the &lt;math&gt;z&lt;/math&gt;'s cancel out, and we are left with &lt;math&gt;3/4&lt;/math&gt;.<br /> Thus, the answer is &lt;math&gt;\mathrm{(B)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> ~ProGameXD<br /> <br /> ==See also==<br /> {{AMC12 box|year=2007|ab=A|num-b=8|num-a=10}}<br /> {{AMC10 box|year=2007|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_9&diff=86544 2007 AMC 12A Problems/Problem 9 2017-07-24T12:27:37Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #9]] and [[2007 AMC 10A Problems/Problem 13|2007 AMC 10A #13]]}}<br /> ==Problem==<br /> Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the [[ratio]] of Yan's distance from his home to his distance from the stadium?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the distance from Yan's initial position to the stadium be &lt;math&gt;a&lt;/math&gt; and the distance from Yan's initial position to home be &lt;math&gt;b&lt;/math&gt;. We are trying to find &lt;math&gt;b/a&lt;/math&gt;, and we have the following identity given by the problem:<br /> <br /> &lt;cmath&gt;\begin{align*}a &amp;= b + \frac{a+b}{7}\\<br /> \frac{6a}{7} &amp;= \frac{8b}{7} \Longrightarrow 6a = 8b\end{align*}&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;b/a = 6/8 = 3/4&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(B)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Another way of solving this problem is by setting the distance between Yan's home and the stadium, thus filling in one variable.<br /> Let us set the distance between the two places to be &lt;math&gt;42z&lt;/math&gt;, where &lt;math&gt;z&lt;/math&gt; is a random measurement (cause life, why not?). The distance to going to his home then riding his bike, which is &lt;math&gt;7&lt;/math&gt; times faster, is equal to him just walking to the stadium. So the equation would be:<br /> Let &lt;math&gt;x=&lt;/math&gt; the distance from Yan's position to his home. <br /> Let &lt;math&gt;42z=&lt;/math&gt; the distance from Yan's home to the stadium.<br /> &lt;math&gt;42-x=x+42/7&lt;/math&gt;<br /> <br /> &lt;math&gt;42-x=x+6&lt;/math&gt;<br /> <br /> &lt;math&gt;36=2x&lt;/math&gt;<br /> <br /> &lt;math&gt;x=18&lt;/math&gt;<br /> <br /> But we're still not done with the question. We know that Yan is &lt;math&gt;18z&lt;/math&gt; from his home, and is &lt;math&gt;42z-18z&lt;/math&gt; or &lt;math&gt;24z&lt;/math&gt; from the stadium.<br /> &lt;math&gt;18z/24z&lt;/math&gt;, the &lt;math&gt;z&lt;/math&gt;'s cancel out, and we are left with &lt;math&gt;3/4&lt;/math&gt;.<br /> Thus, the answer is &lt;math&gt;\mathrm{(B)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> ~ProGameXD<br /> <br /> ==See also==<br /> {{AMC12 box|year=2007|ab=A|num-b=8|num-a=10}}<br /> {{AMC10 box|year=2007|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_15&diff=86542 2007 AMC 10A Problems/Problem 15 2017-07-24T11:39:43Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Four circles of radius &lt;math&gt;1&lt;/math&gt; are each tangent to two sides of a square and externally tangent to a circle of radius &lt;math&gt;2&lt;/math&gt;, as shown. What is the area of the square?<br /> <br /> [[Image:2007 AMC 10A -15 for wiki.png]]<br /> <br /> &lt;math&gt;\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> <br /> The diagonal has length &lt;math&gt;\sqrt{2}+1+2+2+1+\sqrt{2}=6+2\sqrt{2}&lt;/math&gt;. Therefore the sides have length &lt;math&gt;2+3\sqrt{2}&lt;/math&gt;, and the area is<br /> <br /> &lt;cmath&gt;A=(2+3\sqrt{2})^2=4+6\sqrt{2}+6\sqrt{2}+18=22+12\sqrt{2} \Rightarrow \text{(B)}&lt;/cmath&gt;<br /> <br /> === Solution 2 === <br /> <br /> <br /> <br /> Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides &lt;math&gt;3, 3, 3\sqrt{2}&lt;/math&gt;. The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then &lt;cmath&gt; A = (2+3\sqrt{2})^2 = 22 + 12\sqrt{2} \Rightarrow \text{(B)}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2007|ab=A|num-b=14|num-a=16}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_15&diff=86541 2007 AMC 10A Problems/Problem 15 2017-07-24T11:39:34Z <p>Progamexd: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Four circles of radius &lt;math&gt;1&lt;/math&gt; are each tangent to two sides of a square and externally tangent to a circle of radius &lt;math&gt;2&lt;/math&gt;, as shown. What is the area of the square?<br /> <br /> [[Image:2007 AMC 10A -15 for wiki.png]]<br /> <br /> &lt;math&gt;\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> <br /> The diagonal has length &lt;math&gt;\sqrt{2}+1+2+2+1+\sqrt{2}=6+2\sqrt{2}&lt;/math&gt;. Therefore the sides have length &lt;math&gt;2+3\sqrt{2}&lt;/math&gt;, and the area is<br /> <br /> &lt;cmath&gt;A=(2+3\sqrt{2})^2=4+6\sqrt{2}+6\sqrt{2}+18=22+12\sqrt{2} \Rightarrow \text{(B)}&lt;/cmath&gt;<br /> <br /> === Solution 2 === <br /> <br /> ----<br /> <br /> <br /> Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides &lt;math&gt;3, 3, 3\sqrt{2}&lt;/math&gt;. The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then &lt;cmath&gt; A = (2+3\sqrt{2})^2 = 22 + 12\sqrt{2} \Rightarrow \text{(B)}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2007|ab=A|num-b=14|num-a=16}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_9&diff=86514 2007 AMC 12A Problems/Problem 9 2017-07-23T16:19:54Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #9]] and [[2007 AMC 10A Problems/Problem 13|2007 AMC 10A #13]]}}<br /> ==Problem==<br /> Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the [[ratio]] of Yan's distance from his home to his distance from the stadium?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the distance from Yan's initial position to the stadium be &lt;math&gt;a&lt;/math&gt; and the distance from Yan's initial position to home be &lt;math&gt;b&lt;/math&gt;. We are trying to find &lt;math&gt;b/a&lt;/math&gt;, and we have the following identity given by the problem:<br /> <br /> &lt;cmath&gt;\begin{align*}a &amp;= b + \frac{a+b}{7}\\<br /> \frac{6a}{7} &amp;= \frac{8b}{7} \Longrightarrow 6a = 8b\end{align*}&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;b/a = 6/8 = 3/4&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(B)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Another way of solving this problem is by setting the distance between Yan's home and the stadium, thus filling in one variable.<br /> Let us set the distance between the two places to be &lt;math&gt;42z&lt;/math&gt;, where &lt;math&gt;z&lt;/math&gt; is a random measurement (cause life, why not?). The distance to going to his home then riding his bike, which is &lt;math&gt;7&lt;/math&gt; times faster, is equal to him just walking to the stadium. So the equation would be:<br /> Let &lt;math&gt;x=&lt;/math&gt; the distance from Yan's position to his home. <br /> Let &lt;math&gt;42z=&lt;/math&gt; the distance from Yan's home to the stadium.<br /> &lt;math&gt;42-x=x+42/7&lt;/math&gt;<br /> &lt;math&gt;42-x=x+6&lt;/math&gt;<br /> &lt;math&gt;36=2x&lt;/math&gt;<br /> &lt;math&gt;x=18&lt;/math&gt;<br /> But we're still not done with the question. We know that Yan is &lt;math&gt;18z&lt;/math&gt; from his home, and is &lt;math&gt;42z-18z&lt;/math&gt; or &lt;math&gt;24z&lt;/math&gt; from the stadium.<br /> &lt;math&gt;18z/24z&lt;/math&gt;, the &lt;math&gt;z&lt;/math&gt;'s cancel out, and we are left with &lt;math&gt;3/4&lt;/math&gt;.<br /> Thus, the answer is &lt;math&gt;\mathrm{(B)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> ~ProGameXD<br /> <br /> ==See also==<br /> {{AMC12 box|year=2007|ab=A|num-b=8|num-a=10}}<br /> {{AMC10 box|year=2007|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_9&diff=86513 2007 AMC 12A Problems/Problem 9 2017-07-23T16:19:44Z <p>Progamexd: /* Solution */</p> <hr /> <div>{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #9]] and [[2007 AMC 10A Problems/Problem 13|2007 AMC 10A #13]]}}<br /> ==Problem==<br /> Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the [[ratio]] of Yan's distance from his home to his distance from the stadium?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the distance from Yan's initial position to the stadium be &lt;math&gt;a&lt;/math&gt; and the distance from Yan's initial position to home be &lt;math&gt;b&lt;/math&gt;. We are trying to find &lt;math&gt;b/a&lt;/math&gt;, and we have the following identity given by the problem:<br /> <br /> &lt;cmath&gt;\begin{align*}a &amp;= b + \frac{a+b}{7}\\<br /> \frac{6a}{7} &amp;= \frac{8b}{7} \Longrightarrow 6a = 8b\end{align*}&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;b/a = 6/8 = 3/4&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(B)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Another way of solving this problem is by setting the distance between Yan's home and the stadium, thus filling in one variable.<br /> Let us set the distance between the two places to be &lt;math&gt;42z&lt;/math&gt;, where &lt;math&gt;z&lt;/math&gt; is a random measurement (cause life, why not?). The distance to going to his home then riding his bike, which is &lt;math&gt;7&lt;/math&gt; times faster, is equal to him just walking to the stadium. So the equation would be:<br /> Let &lt;math&gt;x=&lt;/math&gt; the distance from Yan's position to his home. <br /> Let &lt;math&gt;42z=&lt;/math&gt; the distance from Yan's home to the stadium.<br /> &lt;math&gt;42-x=x+42/7&lt;/math&gt;<br /> &lt;math&gt;42-x=x+6&lt;/math&gt;<br /> &lt;math&gt;36=2x&lt;/math&gt;<br /> &lt;math&gt;x=18&lt;/math&gt;<br /> But we're still not done with the question. We know that Yan is &lt;math&gt;18z&lt;/math&gt; from his home, and is &lt;math&gt;42z-18z&lt;/math&gt; or &lt;math&gt;24z&lt;/math&gt; from the stadium.<br /> &lt;math&gt;18z/24z&lt;/math&gt;, the &lt;math&gt;z&lt;/math&gt;'s cancel out, and we are left with &lt;math&gt;3/4&lt;/math&gt;.<br /> Thus, the answer is &lt;math&gt;\mathrm{(B)}\ \frac{3}{4}&lt;/math&gt;<br /> ~ProGameXD<br /> <br /> ==See also==<br /> {{AMC12 box|year=2007|ab=A|num-b=8|num-a=10}}<br /> {{AMC10 box|year=2007|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_9&diff=86512 2007 AMC 10A Problems/Problem 9 2017-07-23T16:02:02Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; satisfy the [[equation]]s &lt;math&gt;3^{a} = 81^{b + 2}&lt;/math&gt; and &lt;math&gt;125^{b} = 5^{a - 3}&lt;/math&gt;. What is &lt;math&gt;ab&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\ -60 \qquad \text{(B)}\ -17 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 60&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> &lt;cmath&gt;81^{b+2} = 3^{4(b+2)} = 3^a \Longrightarrow a = 4b+8&lt;/cmath&gt;<br /> <br /> And<br /> <br /> &lt;cmath&gt;125^{b} = 5^{3b} = 5^{a-3} \Longrightarrow a - 3 = 3b&lt;/cmath&gt;<br /> <br /> Substitution gives &lt;math&gt;4b+8 - 3 = 3b \Longrightarrow b = -5&lt;/math&gt;, and solving for &lt;math&gt;a&lt;/math&gt; yields &lt;math&gt;-12&lt;/math&gt;. Thus &lt;math&gt;ab = 60\ \mathrm{(E)}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Simplify equation &lt;math&gt;1&lt;/math&gt; which is &lt;math&gt;3^a=81^b+2&lt;/math&gt;, to &lt;math&gt;3^a=3^4(b+2)&lt;/math&gt;, which equals &lt;math&gt;3^a=3^4b+8&lt;/math&gt;. <br /> <br /> And<br /> <br /> Simplify equation &lt;math&gt;2&lt;/math&gt; which is &lt;math&gt;125^b=5^a-3&lt;/math&gt;, to &lt;math&gt;5^3(b)=5^a-3&lt;/math&gt;, which equals &lt;math&gt;5^3b=5^a-3&lt;/math&gt;.<br /> <br /> Now, eliminate the bases from the simplified equations &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; to arrive at &lt;math&gt;a=4b+8&lt;/math&gt; and &lt;math&gt;3b=a-3&lt;/math&gt;. Rewrite equation &lt;math&gt;2&lt;/math&gt; so that it is in terms of &lt;math&gt;a&lt;/math&gt;. That would be &lt;math&gt;a=3b+3&lt;/math&gt;.<br /> <br /> Since both equations are equal to &lt;math&gt;a&lt;/math&gt;, and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are the same number for both problems, set the equations equal to each other.<br /> &lt;math&gt;4b+8=3b+3&lt;/math&gt;<br /> &lt;math&gt;b=-5&lt;/math&gt;<br /> <br /> Now plug &lt;math&gt;b&lt;/math&gt;, which is &lt;math&gt;(-5)&lt;/math&gt; back into one of the two earlier equations.<br /> &lt;math&gt;4(-5)+8=a&lt;/math&gt;<br /> &lt;math&gt;-20+8=a&lt;/math&gt;<br /> &lt;math&gt;a=-12&lt;/math&gt;<br /> <br /> &lt;math&gt;(-12)(-5)=60&lt;/math&gt;<br /> <br /> Therefore the correct answer is E<br /> <br /> == See also ==<br /> {{AMC10 box|year=2007|ab=A|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_20&diff=86338 2017 AMC 10A Problems/Problem 20 2017-07-11T14:42:47Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; equal the sum of the digits of positive integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;S(1507) = 13&lt;/math&gt;. For a particular positive integer &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S(n) = 1274&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;S(n+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;n \equiv S(n) \pmod{9}&lt;/math&gt;. This can be seen from the fact that &lt;math&gt;\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}&lt;/math&gt;. Thus, if &lt;math&gt;S(n) = 1274&lt;/math&gt;, then &lt;math&gt;n \equiv 5 \pmod{9}&lt;/math&gt;, and thus &lt;math&gt;n+1 \equiv S(n+1) \equiv 6 \pmod{9}&lt;/math&gt;. The only answer choice that is &lt;math&gt;6 \pmod{9}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can find out that the least number of digits the number &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;142&lt;/math&gt;, with &lt;math&gt;141&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;'s and one &lt;math&gt;5&lt;/math&gt;.<br /> By randomly mixing the digits up, we are likely to get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9995999&lt;/math&gt;...&lt;math&gt;9999&lt;/math&gt;.<br /> By adding &lt;math&gt;1&lt;/math&gt; to this number, we get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9996000&lt;/math&gt;...&lt;math&gt;0000&lt;/math&gt;.<br /> Knowing that this number is ONLY divisible by &lt;math&gt;9&lt;/math&gt; when &lt;math&gt;6&lt;/math&gt; is subtracted, we can subtract &lt;math&gt;6&lt;/math&gt; from every available choice, and see if the number is divisible by &lt;math&gt;9&lt;/math&gt; afterwards.<br /> After subtracting &lt;math&gt;6&lt;/math&gt; from every number, we can conclude that &lt;math&gt;1233&lt;/math&gt; (originally &lt;math&gt;1239&lt;/math&gt;) is the only number divisible by &lt;math&gt;9&lt;/math&gt;. <br /> So our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> The number &lt;math&gt;n&lt;/math&gt; can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice.<br /> <br /> If &lt;math&gt;\boxed{\textbf{(A)}\ 1}&lt;/math&gt; is correct, then &lt;math&gt;n&lt;/math&gt; must be some number &lt;math&gt;99999999...9&lt;/math&gt;, because when we add one to &lt;math&gt;99999999...9&lt;/math&gt; we get &lt;math&gt;10000000...00&lt;/math&gt;. Thus, if &lt;math&gt;1&lt;/math&gt; is the correct answer, then the equation &lt;math&gt;9x=1274&lt;/math&gt; must have an integer solution (i.e. &lt;math&gt;1274&lt;/math&gt; must be divisible by &lt;math&gt;9&lt;/math&gt;). But since it does not, &lt;math&gt;1&lt;/math&gt; is not the correct answer.<br /> <br /> If &lt;math&gt;\boxed{\textbf{(B)}\ 3}&lt;/math&gt; is correct, then &lt;math&gt;n&lt;/math&gt; must be some number &lt;math&gt;29999999...9&lt;/math&gt;, because when we add one to &lt;math&gt;29999999...9&lt;/math&gt;, we get &lt;math&gt;30000000...00&lt;/math&gt;. Thus, if &lt;math&gt;2&lt;/math&gt; is the correct answer, then the equation &lt;math&gt;2+9x=1274&lt;/math&gt; must have an integer solution. But since it does not, &lt;math&gt;3&lt;/math&gt; is not the correct answer.<br /> <br /> Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities. Notice that if &lt;math&gt;S(n+1)=N&lt;/math&gt;, then &lt;math&gt;n&lt;/math&gt; must be a number whose initial digits sum to &lt;math&gt;N-1&lt;/math&gt;, and whose other, terminating digits, are all &lt;math&gt;9&lt;/math&gt;. Thus, we can evaluate the three final possibilities by seeing if the equation &lt;math&gt;(N-1)+9x=1274&lt;/math&gt; has an integer solution.<br /> <br /> The equation does not have an integer solution for &lt;math&gt;N=12&lt;/math&gt;, so &lt;math&gt;\boxed{\textbf{(C)}\ 12}&lt;/math&gt; is not correct. However, the equation does have an integer solution for &lt;math&gt;N=1239&lt;/math&gt; (&lt;math&gt;x=4&lt;/math&gt;), so &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt; is the answer.<br /> <br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_17&diff=86240 2003 AMC 10A Problems/Problem 17 2017-07-01T13:28:38Z <p>Progamexd: /* Solution */</p> <hr /> <div>== Problem ==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{3\sqrt{2}}{\pi}\qquad \mathrm{(B) \ } \frac{3\sqrt{3}}{\pi}\qquad \mathrm{(C) \ } \sqrt{3}\qquad \mathrm{(D) \ } \frac{6}{\pi}\qquad \mathrm{(E) \ } \sqrt{3}\pi &lt;/math&gt;<br /> <br /> == Solution ==<br /> Let &lt;math&gt;s&lt;/math&gt; be the length of a side of the equilateral triangle and let &lt;math&gt;r&lt;/math&gt; be the radius of the circle. <br /> <br /> In a circle with a radius &lt;math&gt;r&lt;/math&gt;, the side of an inscribed equilateral triangle is &lt;math&gt;r\sqrt{3}&lt;/math&gt;. <br /> <br /> So &lt;math&gt;s=r\sqrt{3}&lt;/math&gt;. <br /> <br /> The perimeter of the triangle is &lt;math&gt;3s=3r\sqrt{3}&lt;/math&gt;<br /> <br /> The area of the circle is &lt;math&gt;\pi r^{2}&lt;/math&gt;<br /> <br /> So: <br /> &lt;math&gt;\pi r^{2} = 3r\sqrt{3}&lt;/math&gt;<br /> <br /> &lt;math&gt;\pi r=3\sqrt{3}&lt;/math&gt;<br /> <br /> &lt;math&gt;r=\frac{3\sqrt{3}}{\pi} \Rightarrow\boxed{\mathrm{(B)}\ \frac{3\sqrt{3}}{\pi}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_20&diff=85101 2017 AMC 10A Problems/Problem 20 2017-03-31T23:03:34Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; equal the sum of the digits of positive integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;S(1507) = 13&lt;/math&gt;. For a particular positive integer &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S(n) = 1274&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;S(n+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;n \equiv S(n) \pmod{9}&lt;/math&gt;. This can be seen from the fact that &lt;math&gt;\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}&lt;/math&gt;. Thus, if &lt;math&gt;S(n) = 1274&lt;/math&gt;, then &lt;math&gt;n \equiv 5 \pmod{9}&lt;/math&gt;, and thus &lt;math&gt;n+1 \equiv S(n+1) \equiv 6 \pmod{9}&lt;/math&gt;. The only answer choice that is &lt;math&gt;6 \pmod{9}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can find out that the least number of digits the number &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;142&lt;/math&gt;, with &lt;math&gt;141&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;'s and one &lt;math&gt;5&lt;/math&gt;.<br /> By randomly mixing the digits up, we are likely to get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9995999&lt;/math&gt;...&lt;math&gt;9999&lt;/math&gt;.<br /> By adding &lt;math&gt;1&lt;/math&gt; to this number, we get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9996000&lt;/math&gt;...&lt;math&gt;0000&lt;/math&gt;.<br /> Knowing that this number is ONLY divisible by &lt;math&gt;9&lt;/math&gt; when &lt;math&gt;6&lt;/math&gt; is subtracted, we can subtract &lt;math&gt;6&lt;/math&gt; from every available choice, and see if the number is divisible by &lt;math&gt;9&lt;/math&gt; afterwards.<br /> After subtracting &lt;math&gt;6&lt;/math&gt; from every number, we can conclude that &lt;math&gt;1233&lt;/math&gt; (originally &lt;math&gt;1239&lt;/math&gt;) is the only number divisible by &lt;math&gt;9&lt;/math&gt;. <br /> So our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ~ProGameXD<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_20&diff=85100 2017 AMC 10A Problems/Problem 20 2017-03-31T23:03:22Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; equal the sum of the digits of positive integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;S(1507) = 13&lt;/math&gt;. For a particular positive integer &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S(n) = 1274&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;S(n+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;n \equiv S(n) \pmod{9}&lt;/math&gt;. This can be seen from the fact that &lt;math&gt;\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}&lt;/math&gt;. Thus, if &lt;math&gt;S(n) = 1274&lt;/math&gt;, then &lt;math&gt;n \equiv 5 \pmod{9}&lt;/math&gt;, and thus &lt;math&gt;n+1 \equiv S(n+1) \equiv 6 \pmod{9}&lt;/math&gt;. The only answer choice that is &lt;math&gt;6 \pmod{9}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can find out that the least number of digits the number &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;142&lt;/math&gt;, with &lt;math&gt;141&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;'s and one &lt;math&gt;5&lt;/math&gt;.<br /> By randomly mixing the digits up, we are likely to get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9995999&lt;/math&gt;...&lt;math&gt;9999&lt;/math&gt;.<br /> By adding &lt;math&gt;1&lt;/math&gt; to this number, we get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9996000&lt;/math&gt;...&lt;math&gt;0000&lt;/math&gt;.<br /> Knowing that this number is ONLY divisible by &lt;math&gt;9&lt;/math&gt; when &lt;math&gt;6&lt;/math&gt; is subtracted, we can subtract &lt;math&gt;6&lt;/math&gt; from every available choice, and see if the number is divisible by &lt;math&gt;9&lt;/math&gt; afterwards.<br /> After subtracting &lt;math&gt;6&lt;/math&gt; from every number, we can conclude that &lt;math&gt;1233&lt;/math&gt; (originally &lt;math&gt;1239&lt;/math&gt;) is the only number divisible by &lt;math&gt;9&lt;/math&gt;. <br /> So our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> ~ProGameXD<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_20&diff=83406 2017 AMC 10A Problems/Problem 20 2017-02-10T15:31:13Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; equal the sum of the digits of positive integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;S(1507) = 13&lt;/math&gt;. For a particular positive integer &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S(n) = 1274&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;S(n+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;n \equiv S(n) \pmod{9}&lt;/math&gt;. This can be seen from the fact that &lt;math&gt;\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}&lt;/math&gt;. Thus, if &lt;math&gt;S(n) = 1274&lt;/math&gt;, then &lt;math&gt;n \equiv 5 \pmod{9}&lt;/math&gt;, and thus &lt;math&gt;n+1 \equiv S(n+1) \equiv 6 \pmod{9}&lt;/math&gt;. The only answer choice that is &lt;math&gt;6 \pmod{9}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can find out that the least number of digits the number &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;142&lt;/math&gt;, with &lt;math&gt;141&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;'s and one &lt;math&gt;5&lt;/math&gt;.<br /> By randomly mixing the digits up, we are likely to get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9995999&lt;/math&gt;...&lt;math&gt;9999&lt;/math&gt;.<br /> By adding 1 to this number, we get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9996000&lt;/math&gt;...&lt;math&gt;0000&lt;/math&gt;.<br /> We can subtract 6 from every available choice, and see if the number is divisible by &lt;math&gt;9&lt;/math&gt; afterwards.<br /> After subtracting 6 from every number, we can conclude that &lt;math&gt;1233&lt;/math&gt; (originally &lt;math&gt;1239&lt;/math&gt;) is the only number divisible by 9. <br /> So our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_20&diff=83377 2017 AMC 10A Problems/Problem 20 2017-02-09T23:07:45Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; equal the sum of the digits of positive integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;S(1507) = 13&lt;/math&gt;. For a particular positive integer &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S(n) = 1274&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;S(n+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;n \equiv S(n) \pmod{9}&lt;/math&gt;. This can be seen from the fact that &lt;math&gt;\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}&lt;/math&gt;. Thus, if &lt;math&gt;S(n) = 1274&lt;/math&gt;, then &lt;math&gt;n \equiv 5 \pmod{9}&lt;/math&gt;, and thus &lt;math&gt;n+1 \equiv S(n+1) \equiv 6 \pmod{9}&lt;/math&gt;. The only answer choice that is &lt;math&gt;6 \pmod{9}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can find out that the least number of digits the number &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;142&lt;/math&gt;, with &lt;math&gt;141&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;'s and one &lt;math&gt;5&lt;/math&gt;.<br /> By randomly mixing the digits up, we are likely to get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9995999&lt;/math&gt;...&lt;math&gt;9999&lt;/math&gt;.<br /> By adding 1 to this number, we get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9996000&lt;/math&gt;...&lt;math&gt;0000&lt;/math&gt;.<br /> We can subtract 6 from every available choice, and see if the number is divisible by 9 afterwards.<br /> After subtracting 6 from every number, we can conclude that &lt;math&gt;1233&lt;/math&gt; (originally &lt;math&gt;1239&lt;/math&gt;) is the only number divisible by 9. <br /> So our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ~ProGameXD<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_20&diff=83376 2017 AMC 10A Problems/Problem 20 2017-02-09T23:07:32Z <p>Progamexd: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; equal the sum of the digits of positive integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;S(1507) = 13&lt;/math&gt;. For a particular positive integer &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S(n) = 1274&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;S(n+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;n \equiv S(n) \pmod{9}&lt;/math&gt;. This can be seen from the fact that &lt;math&gt;\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}&lt;/math&gt;. Thus, if &lt;math&gt;S(n) = 1274&lt;/math&gt;, then &lt;math&gt;n \equiv 5 \pmod{9}&lt;/math&gt;, and thus &lt;math&gt;n+1 \equiv S(n+1) \equiv 6 \pmod{9}&lt;/math&gt;. The only answer choice that is &lt;math&gt;6 \pmod{9}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can find out that the least number of digits the number &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;142&lt;/math&gt;, with &lt;math&gt;141&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;'s and one &lt;math&gt;5&lt;/math&gt;.<br /> By randomly mixing the digits up, we are likely to get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9995999&lt;/math&gt;...&lt;math&gt;9999&lt;/math&gt;.<br /> By adding 1 to this number, we get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9996000&lt;/math&gt;...&lt;math&gt;0000&lt;/math&gt;.<br /> We can subtract 6 from every available choice, and see if the number is divisible by 9 afterwards.<br /> After subtracting 6 from every number, we can conclude that &lt;math&gt;1233&lt;/math&gt; (originally &lt;math&gt;1239&lt;/math&gt;) is the only number divisible by 9. <br /> So our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 1239}&lt;/math&gt;.<br /> ~ProGameXD<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_23&diff=83375 2017 AMC 10A Problems/Problem 23 2017-02-09T23:05:26Z <p>Progamexd: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> How many triangles with positive area have all their vertices at points &lt;math&gt;(i,j)&lt;/math&gt; in the coordinate plane, where &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;, inclusive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300&lt;/math&gt;<br /> <br /> ==Solution==<br /> There are a total of &lt;math&gt;\binom{25}{3}=2300&lt;/math&gt; sets of three points. However, some of them form degenerate triangles (i.e., they have area of 0) if the three points are collinear. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), which contributes &lt;math&gt;\binom{5}{3} \cdot 12 = 120&lt;/math&gt; degenerate triangles, 4 lines that go through exactly 4 points, which contributes &lt;math&gt;\binom{4}{3} \cdot 4 = 16&lt;/math&gt; degenerate triangles, and 16 lines that go through exactly three points, which contributes &lt;math&gt;\binom{3}{3} \cdot 16 = 16&lt;/math&gt; degenerate triangles. Subtracting these degenerate triangles, we get an answer of &lt;math&gt;2300-120-16-16=2300-152=\boxed{\textbf{(B) }2148}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_23&diff=83374 2017 AMC 10A Problems/Problem 23 2017-02-09T23:05:16Z <p>Progamexd: </p> <hr /> <div>==Problem==<br /> How many triangles with positive area have all their vertices at points &lt;math&gt;(i,j)&lt;/math&gt; in the coordinate plane, where &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;, inclusive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There are a total of &lt;math&gt;\binom{25}{3}=2300&lt;/math&gt; sets of three points. However, some of them form degenerate triangles (i.e., they have area of 0) if the three points are collinear. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), which contributes &lt;math&gt;\binom{5}{3} \cdot 12 = 120&lt;/math&gt; degenerate triangles, 4 lines that go through exactly 4 points, which contributes &lt;math&gt;\binom{4}{3} \cdot 4 = 16&lt;/math&gt; degenerate triangles, and 16 lines that go through exactly three points, which contributes &lt;math&gt;\binom{3}{3} \cdot 16 = 16&lt;/math&gt; degenerate triangles. Subtracting these degenerate triangles, we get an answer of &lt;math&gt;2300-120-16-16=2300-152=\boxed{\textbf{(B) }2148}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_23&diff=83373 2017 AMC 10A Problems/Problem 23 2017-02-09T23:04:43Z <p>Progamexd: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> How many triangles with positive area have all their vertices at points &lt;math&gt;(i,j)&lt;/math&gt; in the coordinate plane, where &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;, inclusive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There are a total of &lt;math&gt;\binom{25}{3}=2300&lt;/math&gt; sets of three points. However, some of them form degenerate triangles (i.e., they have area of 0) if the three points are collinear. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), which contributes &lt;math&gt;\binom{5}{3} \cdot 12 = 120&lt;/math&gt; degenerate triangles, 4 lines that go through exactly 4 points, which contributes &lt;math&gt;\binom{4}{3} \cdot 4 = 16&lt;/math&gt; degenerate triangles, and 16 lines that go through exactly three points, which contributes &lt;math&gt;\binom{3}{3} \cdot 16 = 16&lt;/math&gt; degenerate triangles. Subtracting these degenerate triangles, we get an answer of &lt;math&gt;2300-120-16-16=2300-152=\boxed{\textbf{(B) }2148}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can find out that the least number of digits the number &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;142&lt;/math&gt;, with &lt;math&gt;141&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;'s and one &lt;math&gt;5&lt;/math&gt;.<br /> By randomly mixing the digits up, we are likely to get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9995999&lt;/math&gt;...&lt;math&gt;9999&lt;/math&gt;.<br /> By adding 1 to this number, we get: &lt;math&gt;9999&lt;/math&gt;...&lt;math&gt;9996000&lt;/math&gt;...&lt;math&gt;0000&lt;/math&gt;.<br /> We can subtract 6 from every available choice, and see if the number is divisible by 9 afterwards.<br /> After subtracting 6 from every number, we can conclude that &lt;math&gt;1233&lt;/math&gt; (originally &lt;math&gt;1239&lt;/math&gt;) is the only number divisible by 9. <br /> So our answer is &lt;math&gt;1239&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_23&diff=83372 2017 AMC 10A Problems/Problem 23 2017-02-09T22:59:25Z <p>Progamexd: /* Solution */</p> <hr /> <div>==Problem==<br /> How many triangles with positive area have all their vertices at points &lt;math&gt;(i,j)&lt;/math&gt; in the coordinate plane, where &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;, inclusive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There are a total of &lt;math&gt;\binom{25}{3}=2300&lt;/math&gt; sets of three points. However, some of them form degenerate triangles (i.e., they have area of 0) if the three points are collinear. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), which contributes &lt;math&gt;\binom{5}{3} \cdot 12 = 120&lt;/math&gt; degenerate triangles, 4 lines that go through exactly 4 points, which contributes &lt;math&gt;\binom{4}{3} \cdot 4 = 16&lt;/math&gt; degenerate triangles, and 16 lines that go through exactly three points, which contributes &lt;math&gt;\binom{3}{3} \cdot 16 = 16&lt;/math&gt; degenerate triangles. Subtracting these degenerate triangles, we get an answer of &lt;math&gt;2300-120-16-16=2300-152=\boxed{\textbf{(B) }2148}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can find out that the least number of digits the number &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;142&lt;/math&gt;, with &lt;math&gt;141&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;'s<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=82737 2016 AMC 10A Problems 2017-02-06T23:41:57Z <p>Progamexd: /* Problem 9 */</p> <hr /> <div>==Problem 1==<br /> What is the value of &lt;math&gt;\dfrac{11!-10!}{9!}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> For what value of &lt;math&gt;x&lt;/math&gt; does &lt;math&gt;10^{x}\cdot 100^{2x}=1000^{5}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> For every dollar Ben spent on bagels, David spent &lt;math&gt;25&lt;/math&gt; cents less. Ben paid &lt;math&gt;\$12.50&lt;/math&gt; more than David. How much did they spend in the bagel store together?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> The remainder can be defined for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;y \neq 0&lt;/math&gt; by &lt;cmath&gt;\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor&lt;/cmath&gt;where &lt;math&gt;\left \lfloor \tfrac{x}{y} \right \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;\tfrac{x}{y}&lt;/math&gt;. What is the value of &lt;math&gt;\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Ximena lists the whole numbers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;30&lt;/math&gt; once. Emilio copies Ximena's numbers, replacing each occurrence of the digit &lt;math&gt;2&lt;/math&gt; by the digit &lt;math&gt;1&lt;/math&gt;. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The mean, median, and mode of the &lt;math&gt;7&lt;/math&gt; data values &lt;math&gt;60, 100, x, 40, 50, 200, 90&lt;/math&gt; are all equal to &lt;math&gt;x&lt;/math&gt;. What is the value of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays &lt;math&gt;40&lt;/math&gt; coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> A triangular array of &lt;math&gt;2016&lt;/math&gt; coins has &lt;math&gt;1&lt;/math&gt; coin in the first row, &lt;math&gt;2&lt;/math&gt; coins in the second row, &lt;math&gt;3&lt;/math&gt; coins in the third row, and so on up to &lt;math&gt;N&lt;/math&gt; coins in the &lt;math&gt;N&lt;/math&gt;th row. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is &lt;math&gt;1&lt;/math&gt; foot wide on all four sides. What is the length in feet of the inner rectangle?<br /> &lt;asy&gt;<br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> path rectangle(pair X, pair Y){<br /> return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br /> }<br /> filldraw(rectangle((0,0),(7,5)),gray(0.5));<br /> filldraw(rectangle((1,1),(6,4)),gray(0.75));<br /> filldraw(rectangle((2,2),(5,3)),white);<br /> <br /> label(&quot;$1$&quot;,(0.5,2.5));<br /> draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br /> draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(1.5,2.5));<br /> draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br /> draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(4.5,2.5));<br /> draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br /> draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(4.1,1.5));<br /> draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br /> draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(3.7,0.5));<br /> draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br /> draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br /> &lt;/asy&gt;<br /> <br /> &lt;cmath&gt;\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8&lt;/cmath&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> What is the area of the shaded region of the given &lt;math&gt;8 \times 5&lt;/math&gt; rectangle?<br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$7$&quot;,(9/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> label(&quot;$4$&quot;,(8,3),dir(0));<br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$7$&quot;,(7/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$4$&quot;,(0,2),dir(180));<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Three distinct integers are selected at random between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2016&lt;/math&gt;, inclusive. Which of the following is a correct statement about the probability &lt;math&gt;p&lt;/math&gt; that the product of the three integers is odd?<br /> <br /> &lt;math&gt;\textbf{(A)}\ p&lt;\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}&lt;p&lt;\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p&gt;\dfrac{1}{3}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Five friends sat in a movie theater in a row containing &lt;math&gt;5&lt;/math&gt; seats, numbered &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;5&lt;/math&gt; from left to right. (The directions &quot;left&quot; and &quot;right&quot; are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> How many ways are there to write &lt;math&gt;2016&lt;/math&gt; as the sum of twos and threes, ignoring order? (For example, &lt;math&gt;1008\cdot 2 + 0\cdot 3&lt;/math&gt; and &lt;math&gt;402\cdot 2 + 404\cdot 3&lt;/math&gt; are two such ways.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Seven cookies of radius &lt;math&gt;1&lt;/math&gt; inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br /> <br /> &lt;asy&gt;<br /> draw(circle((0,0),3));<br /> draw(circle((0,0),1));<br /> draw(circle((1,sqrt(3)),1));<br /> draw(circle((-1,sqrt(3)),1)); <br /> draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br /> draw(circle((2,0),1)); draw(circle((-2,0),1)); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> A triangle with vertices &lt;math&gt;A(0, 2)&lt;/math&gt;, &lt;math&gt;B(-3, 2)&lt;/math&gt;, and &lt;math&gt;C(-3, 0)&lt;/math&gt; is reflected about the &lt;math&gt;x&lt;/math&gt;-axis, then the image &lt;math&gt;\triangle A'B'C'&lt;/math&gt; is rotated counterclockwise about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt; to produce &lt;math&gt;\triangle A''B''C''&lt;/math&gt;. Which of the following transformations will return &lt;math&gt;\triangle A''B''C''&lt;/math&gt; to &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; counterclockwise rotation about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; clockwise rotation about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; reflection about the &lt;math&gt;x&lt;/math&gt;-axis <br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; reflection about the line &lt;math&gt;y = x&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; reflection about the &lt;math&gt;y&lt;/math&gt;-axis.<br /> <br /> [[2016 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Let &lt;math&gt;N&lt;/math&gt; be a positive multiple of &lt;math&gt;5&lt;/math&gt;. One red ball and &lt;math&gt;N&lt;/math&gt; green balls are arranged in a line in random order. Let &lt;math&gt;P(N)&lt;/math&gt; be the probability that at least &lt;math&gt;\tfrac{3}{5}&lt;/math&gt; of the green balls are on the same side of the red ball. Observe that &lt;math&gt;P(5)=1&lt;/math&gt; and that &lt;math&gt;P(N)&lt;/math&gt; approaches &lt;math&gt;\tfrac{4}{5}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; grows large. What is the sum of the digits of the least value of &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;P(N) &lt; \tfrac{321}{400}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Each vertex of a cube is to be labeled with an integer &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;8&lt;/math&gt;, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> In rectangle &lt;math&gt;ABCD,&lt;/math&gt; &lt;math&gt;AB=6&lt;/math&gt; and &lt;math&gt;BC=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, and point &lt;math&gt;F&lt;/math&gt; between &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are such that &lt;math&gt;BE=EF=FC&lt;/math&gt;. Segments &lt;math&gt;\overline{AE}&lt;/math&gt; and &lt;math&gt;\overline{AF}&lt;/math&gt; intersect &lt;math&gt;\overline{BD}&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. The ratio &lt;math&gt;BP:PQ:QD&lt;/math&gt; can be written as &lt;math&gt;r:s:t&lt;/math&gt; where the greatest common factor of &lt;math&gt;r,s&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; is 1. What is &lt;math&gt;r+s+t&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20&lt;/math&gt;<br /> <br /> <br /> [[2016 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> For some particular value of &lt;math&gt;N&lt;/math&gt;, when &lt;math&gt;(a+b+c+d+1)^N&lt;/math&gt; is expanded and like terms are combined, the resulting expression contains exactly &lt;math&gt;1001&lt;/math&gt; terms that include all four variables &lt;math&gt;a, b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;, each to some positive power. What is &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 20|Solution]]<br /> ==Problem 21==<br /> Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> A binary operation &lt;math&gt;\diamondsuit&lt;/math&gt; has the properties that &lt;math&gt;a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c&lt;/math&gt; and that &lt;math&gt;a\,\diamondsuit \,a=1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. (Here &lt;math&gt;\cdot&lt;/math&gt; represents multiplication). The solution to the equation &lt;math&gt;2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100&lt;/math&gt; can be written as &lt;math&gt;\tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p+q?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A quadrilateral is inscribed in a circle of radius &lt;math&gt;200\sqrt{2}&lt;/math&gt;. Three of the sides of this quadrilateral have length &lt;math&gt;200&lt;/math&gt;. What is the length of the fourth side?<br /> <br /> &lt;math&gt;\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2016|ab=A|before=[[2015 AMC 10B Problems]]|after=[[2016 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2016 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_4&diff=82733 2009 AMC 10A Problems/Problem 4 2017-02-06T21:43:45Z <p>Progamexd: /* See also */</p> <hr /> <div>==Problem==<br /> Eric plans to compete in a triathlon. He can average &lt;math&gt;2&lt;/math&gt; miles per hour in the &lt;math&gt;\frac{1}{4}&lt;/math&gt;-mile swim and &lt;math&gt;6&lt;/math&gt; miles per hour in the &lt;math&gt;3&lt;/math&gt;-mile run. His goal is to finish the triathlon in &lt;math&gt;2&lt;/math&gt; hours. To accomplish his goal what must his average speed in miles per hour, be for the &lt;math&gt;15&lt;/math&gt;-mile bicycle ride?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ \frac{120}{11}<br /> \qquad<br /> \mathrm{(B)}\ 11<br /> \qquad<br /> \mathrm{(C)}\ \frac{56}{5}<br /> \qquad<br /> \mathrm{(D)}\ \frac{45}{4}<br /> \qquad<br /> \mathrm{(E)}\ 12<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> Since &lt;math&gt;d=rt&lt;/math&gt;, Eric takes &lt;math&gt;\frac{\frac{1}{4}}{2}=\frac{1}{8}&lt;/math&gt; hours for the swim. Then, he takes &lt;math&gt;\frac{3}{6}=\frac{1}{2}&lt;/math&gt; hours for the run. So he needs to take &lt;math&gt;2-\frac{5}{8}=\frac{11}{8}&lt;/math&gt; hours for the &lt;math&gt;15&lt;/math&gt; mile run. This is &lt;math&gt;\frac{15}{\frac{11}{8}}=\frac{120}{11} \frac{\text{miles}}{\text{hour}}&lt;/math&gt;<br /> <br /> &lt;math&gt;\longrightarrow \fbox{A}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_4&diff=82732 2009 AMC 10A Problems/Problem 4 2017-02-06T21:43:10Z <p>Progamexd: /* See also */</p> <hr /> <div>==Problem==<br /> Eric plans to compete in a triathlon. He can average &lt;math&gt;2&lt;/math&gt; miles per hour in the &lt;math&gt;\frac{1}{4}&lt;/math&gt;-mile swim and &lt;math&gt;6&lt;/math&gt; miles per hour in the &lt;math&gt;3&lt;/math&gt;-mile run. His goal is to finish the triathlon in &lt;math&gt;2&lt;/math&gt; hours. To accomplish his goal what must his average speed in miles per hour, be for the &lt;math&gt;15&lt;/math&gt;-mile bicycle ride?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ \frac{120}{11}<br /> \qquad<br /> \mathrm{(B)}\ 11<br /> \qquad<br /> \mathrm{(C)}\ \frac{56}{5}<br /> \qquad<br /> \mathrm{(D)}\ \frac{45}{4}<br /> \qquad<br /> \mathrm{(E)}\ 12<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> Since &lt;math&gt;d=rt&lt;/math&gt;, Eric takes &lt;math&gt;\frac{\frac{1}{4}}{2}=\frac{1}{8}&lt;/math&gt; hours for the swim. Then, he takes &lt;math&gt;\frac{3}{6}=\frac{1}{2}&lt;/math&gt; hours for the run. So he needs to take &lt;math&gt;2-\frac{5}{8}=\frac{11}{8}&lt;/math&gt; hours for the &lt;math&gt;15&lt;/math&gt; mile run. This is &lt;math&gt;\frac{15}{\frac{11}{8}}=\frac{120}{11} \frac{\text{miles}}{\text{hour}}&lt;/math&gt;<br /> <br /> &lt;math&gt;\longrightarrow \fbox{A}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|nu-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Progamexd https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_25&diff=80401 2016 AMC 10B Problems/Problem 25 2016-09-27T00:31:06Z <p>Progamexd: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)&lt;/math&gt;, where &lt;math&gt;\lfloor r \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;r&lt;/math&gt;. How many distinct values does &lt;math&gt;f(x)&lt;/math&gt; assume for &lt;math&gt;x \ge 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}&lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Progamexd